Please explain (1 Viewer)

[Naylyn]

This from Fitzpatricks New Senior Mathematics exercise 35d, I don't fully understand the answer, please help.

15) A light rod AB of length "l" is freely pivoted at its end A which is fixed and carries a mass M at B. The rod is kept in a horizontal position by means of a string joining B and the point C which is distant "h" vertically above A. Find the magnitude of the force in AB when AB is rotating about a vertical aixs through A at "n" revolutions per second.

This is what I did:
Defining T as the tension of the string and x as the angle between AC and BC

If AB rotates at "n" revolutions per second, then the angular velocity,
w = 2pi*n radians per second

Mg=Tcosx
T=Mg/cosx --(1)
tanx=l/h
Taking Tsinx as the centre seeking force and mrw^2 as the force outwards

If Tsinx < mrw^2
Tsinx = mrw^2 - F (where F is the force in AB)
Then
F = Mrw^2 -Tsinx
sub in (1) and values for w and r
F = Ml*4pi^2*n^2 - Mgtanx
F = Ml*4pi^2*n^2 - Mgl/h
F = Ml(4pi^2*n^2 - g/h)

If Tsinx > mrw^2
Tsinx = mrw^2 + F
Then
F = Tsinx - Mrw^2
sub in (1) and values for w and r
F = Mgtanx - Ml*4pi^2*n^2
F = Mgl/h - Ml*4pi^2*n^2
F = Ml(g/h - 4pi^2*n^2)

The answers give only the first one [F = Ml(4pi^2*n^2 - g/h)] and not the second why is the second answer wrong?

 

[Riviet]

From a quick glance, it looks like 2 cases aren't needed since there are no conditions with the tension (T). I think you just had to find F in terms of the constants like g and m (which you have done).

 

[Naylyn]

I think I have it.
If the force outward exceedes the centre seeking force ie
Tsinx < mrw^2, then there will be no tension in the string and so they can't be equated that way.

Edit: This doesn't work, because it is when Tsinx < mrw^2 that i got the right answer... I copied my working out wrong, it's right now (I hope).

 

[cyl123]

This from Fitzpatricks New Senior Mathematics exercise 35d, I don't fully understand the answer, please help.

15) A light rod AB of length "l" is freely pivoted at its end A which is fixed and carries a mass M at B. The rod is kept in a horizontal position by means of a string joining B and the point C which is distant "h" vertically above A. Find the magnitude of the force in AB when AB is rotating about a vertical aixs through A at "n" revolutions per second.

This is what I did:
Defining T as the tension of the string and x as the angle between AC and BC

If AB rotates at "n" revolutions per second, then the angular velocity,
w = 2pi*n radians per second

Mg=Tcosx
T=Mg/cosx --(1)
tanx=l/h
Taking Tsinx as the centre seeking force and mrw^2 as the force outwards

If Tsinx < mrw^2
Tsinx = mrw^2 - F (where F is the force in AB)
Then
F = Mrw^2 -Tsinx
sub in (1) and values for w and r
F = Ml*4pi^2*n^2 - Mgtanx
F = Ml*4pi^2*n^2 - Mgl/h
F = Ml(4pi^2*n^2 - g/h)

If Tsinx > mrw^2
Tsinx = mrw^2 + F
Then
F = Tsinx - Mrw^2
sub in (1) and values for w and r
F = Mgtanx - Ml*4pi^2*n^2
F = Mgl/h - Ml*4pi^2*n^2
F = Ml(g/h - 4pi^2*n^2)

The answers give only the first one [F = Ml(4pi^2*n^2 - g/h)] and not the second why is the second answer wrong?

If B keeps moving in a circle, the net force of the 2 strings on the particle must be equal to the centripetal force, which is mrw^2, or else it wont rotate in a circle.

As tension on AB is acting in same direction as tension in AC, therefore you only take the case where you add the 2 together ie. your first case
So Tsinx + F = mrw^2
Tsinx = mrw^2 -F which is what you had in the first place

If you had Tsinx = mrw^2 + F, that means that F is acting towards B, so multiply it by -1 to get F where F is acting towards A, which is the tension of AB.

Correct me if im wrong

(By the way, the question only asks for magnitude of force, and magnitude of your second answer is the same as the first, but it just acts in the opposite direction)

 

[Naylyn]

If you had Tsinx = mrw^2 + F, that means that F is acting towards B, so multiply it by -1 to get F where F is acting towards A, which is the tension of AB.

Yes! That explains it perfectly. Thankyou so much.